Answer : Option D
Explanation :
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Solution 1
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Let the sum be Rs.x
$MF#%\begin{align}&\text{Amount after 2 years on Rs.x at 4% per annum when interest is compounded annually }\\\\&=\text{x}\left(1 + \dfrac{4}{100}\right)^2 = \text{x}\left(\dfrac{104}{100}\right)^2\\\\&\text{Compound Interest = }\text{x}\left(\dfrac{104}{100}\right)^2 - x \\\\&= x\left[\left(\dfrac{104}{100}\right)^2 - 1\right] = x\left[\left(\dfrac{26}{25}\right)^2 - 1\right] = x\left[\dfrac{676}{625} - 1\right] = x\left[\dfrac{51}{625} \right] = \dfrac{51x}{625}\\\\\\\\ &\text{Simple Interest = }\dfrac{\text{PRT}}{100} = \dfrac{x \times 4 \times 2}{100} = \dfrac{2x}{25}\end{align}$MF#%
Given that difference between compound interest and simple interest is Rs.1
$MF#%\begin{align}&\Rightarrow \dfrac{51x}{625} - \dfrac{2x}{25} = 1\\\\ &\Rightarrow \dfrac{51x - 50x}{625} = 1\\\\ &\Rightarrow x = 625\end{align}$MF#%
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Solution 2
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The difference between compound interest and simple interest on Rs. P for 2 years at R% per annum
$MF#%= \text{P}\left(\dfrac{\text{R}}{100}\right)^2$MF#%
Let the sum be Rs. x.
C.I. = x ( 1 + ( 10 /100 ))^2 - x = 21x / 100
S.I. = (( x * 10 * 2) / 100) = x / 5
(C.I) - (S.I) = ((21x / 100 ) - (x / 5 )) = x / 100
( x / 100 ) = 632
x = 63100
Hence, the sum is Rs.63,100.
Answer : Option A
Explanation :
This means that, simple Interest on Rs.400 for 1 year = 420 - 400 = 20
$MF#%\text{Rate = }\dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 20}{400 \times 1}= 5\%$MF#%
Rs.400 is the interest on the sum for 1st year
$MF#%\text{Hence, sum = }\dfrac{100 \times \text{SI}}{\text{RT}}= \dfrac{100 \times 400}{5 \times 1} = \text{Rs. 8000}$MF#%
Answer : Option D
Explanation :
$MF#%1386 = \text{P}\left(1 + \dfrac{5}{100}\right)\left(1 + \dfrac{10}{100}\right)\left(1 + \dfrac{20}{100}\right)\\\\ 1386 = \text{P}\left(\dfrac{21}{20}\right)\left(\dfrac{11}{10}\right)\left(\dfrac{6}{5}\right)\\\\ \text{P} = \dfrac{1386 \times 20 \times 10 \times 5}{21 \times 11 \times 6} = \dfrac{66\times 20 \times 10 \times 5}{11 \times 6} = 20 \times 10 \times 5 = \text{Rs. 1000}$MF#%
i.e., the sum is Rs.1000
C.I. when interest
compounded yearly = Rs. 
5000 x 
1 + 4 
x 1 + 
x 4 
100 100 = Rs. 
5000 x 26 x 51 
25 50 = Rs. 5304.
C.I. when interest is
compounded half-yearly = Rs. 5000 x 1 + 2 3 100 = Rs. 5000 x 51 x 51 x 51 50 50 50 = Rs. 5306.04
Difference = Rs. (5306.04 - 5304) = Rs. 2.04
Answer : Option D
Explanation :
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Solution 1
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Let the sum be Rs.x
$MF#%\begin{align}&\text{Amount after 2 years on Rs.x at 4% per annum when interest is compounded annually }\\\\&=\text{x}\left(1 + \dfrac{4}{100}\right)^2 = \text{x}\left(\dfrac{104}{100}\right)^2\\\\&\text{Compound Interest = }\text{x}\left(\dfrac{104}{100}\right)^2 - x \\\\&= x\left[\left(\dfrac{104}{100}\right)^2 - 1\right] = x\left[\left(\dfrac{26}{25}\right)^2 - 1\right] = x\left[\dfrac{676}{625} - 1\right] = x\left[\dfrac{51}{625} \right] = \dfrac{51x}{625}\\\\\\\\ &\text{Simple Interest = }\dfrac{\text{PRT}}{100} = \dfrac{x \times 4 \times 2}{100} = \dfrac{2x}{25}\end{align}$MF#%
Given that difference between compound interest and simple interest is Rs.1
$MF#%\begin{align}&\Rightarrow \dfrac{51x}{625} - \dfrac{2x}{25} = 1\\\\ &\Rightarrow \dfrac{51x - 50x}{625} = 1\\\\ &\Rightarrow x = 625\end{align}$MF#%
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Solution 2
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The difference between compound interest and simple interest on Rs. P for 2 years at R% per annum
$MF#%= \text{P}\left(\dfrac{\text{R}}{100}\right)^2$MF#%
Let the sum be Rs. x.
C.I. = x ( 1 + ( 10 /100 ))^2 - x = 21x / 100
S.I. = (( x * 10 * 2) / 100) = x / 5
(C.I) - (S.I) = ((21x / 100 ) - (x / 5 )) = x / 100
( x / 100 ) = 632
x = 63100
Hence, the sum is Rs.63,100.
Answer : Option A
Explanation :
This means that, simple Interest on Rs.400 for 1 year = 420 - 400 = 20
$MF#%\text{Rate = }\dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 20}{400 \times 1}= 5\%$MF#%
Rs.400 is the interest on the sum for 1st year
$MF#%\text{Hence, sum = }\dfrac{100 \times \text{SI}}{\text{RT}}= \dfrac{100 \times 400}{5 \times 1} = \text{Rs. 8000}$MF#%
Answer : Option C
Explanation :
Let the sum be P
The sum P becomes 3P in 4 years on compound interest
$MF#%3\text{P} = \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^4\\\\ \Rightarrow 3 = \left(1 + \dfrac{\text{R}}{100}\right)^4$MF#%
Let the sum P becomes 81P in n years
$MF#%81\text{P} = \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^n\\\\ \Rightarrow 81 = \left(1 + \dfrac{\text{R}}{100}\right)^n \\\\ \Rightarrow (3)^4 = \left(1 + \dfrac{\text{R}}{100}\right)^n \\\\ \Rightarrow \left(\left(1 + \dfrac{\text{R}}{100}\right)^4\right)^4 = \left(1 + \dfrac{\text{R}}{100}\right)^\text{n} \\\\ \Rightarrow \left(1 + \dfrac{\text{R}}{100}\right)^{16}= \left(1 + \dfrac{\text{R}}{100}\right)^\text{n} \\\\ \text{n} = 16$MF#%
i.e, the sum will become 81 times in 16 years
